Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 5} = \dfrac{-x + 30}{x - 5}$
Multiply both sides by $x - 5$ $ \dfrac{x^2}{x - 5} (x - 5) = \dfrac{-x + 30}{x - 5} (x - 5)$ $ x^2 = -x + 30$ Subtract $-x + 30$ from both sides: $ x^2 - (-x + 30) = -x + 30 - (-x + 30)$ $ x^2 + x - 30 = 0$ Factor the expression: $ (x - 5)(x + 6) = 0$ Therefore $x = 5$ or $x = -6$ At $x = 5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 5$, it is an extraneous solution.